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Surds and Indices Practice Test 1

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Question : 8 of 40

Marks: +1, -0

\frac{2^{n+4}-2\times 2^n}{2\times 2^{n+3}}+2^{-3}

is simplified to:

1
$\left(-2^{n+1}+\frac{1}{8}\right)$
$\left(\frac{9}{8}-2^n\right)$
$2^{n+1}$

Solution:

Concept:

Use the laws of indices to combine powers of the same base and simplify.

Explanation:

Given expression:

\frac{2^{n+4}-2\times 2^n}{2\times 2^{n+3}}+2^{-3}

.

Rewrite

2\times 2^n=2^{n+1}

and

2\times 2^{n+3}=2^{n+4}

.

So the expression becomes

\frac{2^{n+4}-2^{n+1}}{2^{n+4}}+\frac{1}{8}

.

Split the fraction:

1-\frac{2^{n+1}}{2^{n+4}}+\frac{1}{8}

.

Simplify

\frac{2^{n+1}}{2^{n+4}}=2^{-3}=\frac{1}{8}

.

Thus,

1-\frac{1}{8}+\frac{1}{8}=1

.

Answer:

1

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